Question

what is the derivative of: (sin(3x))^2 ?

Answer 1

∫sin²(3x)dx

Let u = 3x and du = 3dx, then...

∫sin²(3x)dx = (1/3)∫sin²udu

sin²u = (1- cos2u)/2, trigonometric identity...

(1/3)∫sin²udu = (1/3)∫[(1- cos2u)/2]du

= (1/6)[∫(1)du - ∫(cos2u)du]

=(1/6)[u - (1/2)sin2u] + c

=(1/6)u - (1/12)sin2u + c

As u = 3x, then...

=(1/6)(3x) - (1/12)sin2(3x) + c

=(1/2)x - (1/12)sin6x + c

=D

Answer 2

... I still don't understand...

Answer 3

@piglet9
what am I doing wrong if I do
deriv(sin(3x))^2 = 2(sin(3x))*(cos(3x))(3) = 6(sin(3x))(cos(3x))

Answer 4

did I just do the integral... LOL

Answer 5

yeah... I don't think I've learned integrals...

Answer 6

You're correct:

2sin(3x)∗cos(3x)∗3

=6∗sin(3x)cos(3x)

=3(2sin(3x)cos(3x))

=3(sin(2∗3x))

=3sin(6x)

Answer 7

Hope that helps :)

Answer 8

thanks :)